Partial derivative of an inner product and a linear transformation Let $u(x)$ be a self adjoint operator on $R^n$, and $\langle\underline{\ },\underline{\ }\rangle$ the usual dot product. 
I have to show $f(x)=\langle x,u(x)\rangle$ is differentiable over all of $R^n$ and calculate the partial derivatives.
I'm having trouble finding the partial derivative. My first idea was to write the dot product as in
$$\sum\limits_{i=1}^{n}\langle e_1,u(x)\rangle \langle e_i,x\rangle$$
But I'm having trouble going from there. Any suggestions?
 A: Note that (formal) self-adjointness implies linearity: $$\langle u(cx),y\rangle = \langle cx,u(y)\rangle = c\langle x,u(y)\rangle = c\langle u(x),y\rangle = \langle cu(x),y\rangle$$
and
$$ \langle u(x+y),z\rangle = \langle x+y,u(z)\rangle = \langle x,u(z)\rangle + \langle y,u(z)\rangle = \langle u(x),z\rangle + \langle u(y),z\rangle = \langle u(x)+u(y),z\rangle. $$
This holds for all $x,y,z\in\mathbb{R}^n$ and all $c\in\mathbb{R}$ so by nondegenerateness of the metric, we have that $u(cx) = cu(x)$ and $u(x+y) = u(x)+u(y)$ for all $x,y\in \mathbb{R}^n$ and all $c\in\mathbb{R}$.
Now pass to coordinates. As a linear operator, once we have imposed a basis $(e_1,e_2,\ldots,e_n)$ on $\mathbb{R}^n$, we have coordinates for $u$: $u_{ij} = \langle u(e_i),e_j\rangle$ and the operation of $u$ can be represented by vector multiplication: $$u(\sum x_ie_i) = \sum x_iu(e_i) = \sum x_iu_{ij}e_j.$$
Therefore the function $f$ may be written in coordinates as:
$$ f(x) = \langle x,u(x)\rangle = \sum_j x_ju(x)_j = \sum_{ij} x_jx_iu_{ij}. $$
Ah! We see that $f$ is, in fact, a homogeneous quadratic polynomial. The derivatives of $f$ are now a straightforward computation (bearing in mind that $\frac{\partial}{\partial x_i}x_j = \delta_{ij}$, the Kronecker delta).
